Molecular Basis of Inheritance — Questions & Answers

Sample questions from Molecular Basis of Inheritance

1. Q1. A very low, basal level of lac operon expression is necessary at all times in E. coli, even in the absence of lactose. This is because:

   • A.The repressor is unstable and degrades constantly
   • B.Some permease must be present for lactose to enter the cell and act as inducer in the first place✓
   • C.Glucose is needed to activate the operator
   • D.The i gene is transcribed only when lactose is present
   Solution

   Lactose can act as an inducer only if it enters the cell, and its transport needs permease, a product of the operon itself. Hence a very low level of expression of the lac operon must always be present so that lactose can enter, inactivate the repressor and switch on full expression.

2. Q2. In a paternity dispute, DNA fingerprinting can establish biological parentage because:

   • A.A child's VNTR bands are entirely random and unrelated to either parent
   • B.A child inherits half of its VNTR/STR alleles from the biological father and half from the biological mother, so every band in the child should be traceable to one or the other biological parent✓
   • C.Only mothers contribute to a child's DNA fingerprint
   • D.DNA fingerprinting cannot be used for paternity testing
   Solution

   Since VNTR/STR alleles are inherited in a Mendelian fashion (one set from each biological parent), every band present in the child's DNA fingerprint should be traceable to a corresponding band in either the true biological mother or father, allowing parentage to be confirmed or excluded.

3. Q3. A culture is grown so that its DNA contains 5-bromouracil, a heavy thymine analogue, in both strands and is then shifted to normal medium for exactly one round of replication. On density-gradient centrifugation the expected result is:

   • A.One heavy band and one light band in equal amounts
   • B.One heavy band only
   • C.A single band of intermediate density✓
   • D.One light band only
   Solution

   Semiconservative replication gives every daughter duplex one heavy parental strand and one newly made light strand, so all molecules have the same intermediate density and form a single hybrid band. Two separate heavy and light bands would be the prediction of conservative replication.

4. Q4. Griffith's transformation experiments with Streptococcus pneumoniae demonstrated that heat-killed virulent (S-strain) bacteria could transform non-virulent (R-strain) bacteria into virulent forms. This result primarily suggested that:

   • A.Bacteria cannot exchange genetic material
   • B.A heritable "transforming principle" from the dead S-strain could be transferred to and permanently alter the R-strain✓
   • C.Heat has no effect on bacterial DNA
   • D.Only living bacteria can cause disease
   Solution

   Griffith's experiment showed that some heritable substance ("transforming principle") from the dead virulent bacteria was taken up by living non-virulent bacteria, permanently converting them to virulent, heritable forms — though Griffith himself did not identify the chemical nature of this substance.

5. Q5. DNA polymerase in E. coli polymerises about 2,000 base pairs per second. If the 4.6 x 10^6 bp E. coli chromosome were replicated by a single replication fork moving in one direction only, replication would take approximately:

   • A.About 2 minutes
   • B.About 18 minutes
   • C.About 38 minutes✓
   • D.About 4 hours
   Solution

   Time = 4.6 x 10^6 bp / 2,000 bp per second = 2,300 seconds, roughly 38 minutes. The actual E. coli replication time (about 18 minutes) is shorter because replication starts at the origin and proceeds bidirectionally with two forks.

Other Class 12 Biology chapters